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Cauchy-Schwarz Problem

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that ([tex]\sum_{j=1}^n[/tex]ajbj)2 <= ([tex]\sum_{j=1}^n[/tex]jaj2)([tex]\sum_{j=1}^n[/tex](1/j)bj2)


    2. Relevant equations
    Cauchy-Schwarz Inequality: |<u, v>| <= ||u||*||v||


    3. The attempt at a solution
    If I let u = [tex]\sum_{j=1}^n[/tex]jaj2 and v = [tex]\sum_{j=1}^n[/tex](1/j)bj2, then I have ||u|| = sqrt(<u, u>) and ||v|| = sqrt(<v, v>). Not really sure where to go from here. Any ideas? Thanks!
     
  2. jcsd
  3. Mar 24, 2008 #2

    Vid

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    The other form is more helpful in this problem:
    <u,v>^2 = <u,u><v,v>

    Now, you just have to choose u and v to get the required identity. You know that <u,v> needs to equal to Sum[a_j*b_j]. You also know that the j-th term in the series on the right side is equal to the j-th component of u or v squared.

    Can you finish it from here?
     
  4. Mar 24, 2008 #3
    First of all, thanks for the help!

    Don't you mean <u,v>2 <= <u,u><v,v> ? Also, I'm not sure what you mean by your last sentence. "The series on the right side" is referring to which series? The product of the two series? Anyway, how do I know that? Sorry for these possibly stupid questions, and thanks again for your help.
     
  5. Mar 24, 2008 #4

    Vid

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    Yea, that should be less than or equal to.

    I meant that one series was for u and the other for v. With the standard dot product <u,u> = Sum[(u_j)^2].

    So Sum[u_j^2] = Sum[j*a_j^2]
    and Sum[v_j^2] = Sum[(1/j)*b_j^2]

    What does this tell us about what u and v must be equal to?
     
  6. Mar 24, 2008 #5
    I don't think I'm seeing it. u_j = sqrt(j)*a_j and u = Sum[u_j] = Sum[sqrt(j)*a_j] ?
    v_j = (1/sqrt(j))*b_j and v = Sum[v_j] = Sum[(1/sqrt(j))*b_j] ?

    Is this correct? I don't really see how this helps.
     
  7. Mar 24, 2008 #6
    the cs inequality says <p, q> <= |p||q|, here | | denotes the norm

    so (<p, q>)^2 <= |p|^2|q|^2


    Set p = (a_1/sqrt(1),..., a_n/sqrt(n)), q= (sqrt(1)*b_1, 2*b_2,..., sqrt(n)*b_n)


    so your sum on the left = (<p, q>)^2 <= |p|^2|q|^2 = the stuff you have on the right

    just verify this
     
  8. Mar 25, 2008 #7
    Ok, just to make sure I got it:

    Cauchy Schwarz says |<u, v>| <= ||u||*||v||. This implies <u, v>2 <= <u, u><u, v>. Let <u, u> = [tex]\sum[/tex]uj2 = [tex]\sum[/tex]jaj2 and let <v, v> = [tex]\sum[/tex]vj2 = [tex]\sum[/tex](1/j)bj2.

    Then we have u = [tex]\sum[/tex]uj = [tex]\sum[/tex]sqrt(j)*aj and v = [tex]\sum[/tex]vj = [tex]\sum[/tex](1/sqrt(j))*bj.

    Then we have <u, v> = [tex]\sum[/tex]ujvj = ... = [tex]\sum[/tex]ajbj. Then square both sides to get <u, v>2 = ([tex]\sum[/tex]ajbj)2.

    Then after plugging in, this completes the proof. Thanks again for all your help guys!
     
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